SOLUTION: The formula P=0.64x^2-0.045x+1 models the approximate population P, in thousands, for a species of fish in a local pond, x years after 1997. During what year will the population re

Algebra ->  Expressions-with-variables -> SOLUTION: The formula P=0.64x^2-0.045x+1 models the approximate population P, in thousands, for a species of fish in a local pond, x years after 1997. During what year will the population re      Log On


   

Question 87064This question is from textbook
: The formula P=0.64x^2-0.045x+1 models the approximate population P, in thousands, for a species of fish in a local pond, x years after 1997. During what year will the population reach 64550 fish? This question is from textbook

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
the formula is +P+=+0.64x%5E2+-+0.045x+%2B+1+

Find the year (x) after which th epopulation is 64550 (P). Now P in the formula is in thousands --> P=64.55

So, +64.55+=+0.64x%5E2+-+0.045x+%2B+1+

Multiply everything by 1000 to get rid of decimals: +64550+=+640x%5E2+-+45x+%2B+1000+
+640x%5E2+-+45x+%2B+1000+=+64550+
+640x%5E2+-+45x+-+63550+=+0+ and simplify by factor of 5
+128x%5E2+-+9x+-+12710+=+0+

I don't think i want to try factorising this so straight to formula:

+x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29+%29%2F%282a%29+

+x+=+%289+%2B-+sqrt%2881+%2B+6507520%29+%29%2F%28256%29+
+x+=+%289+%2B-+sqrt%286507601%29+%29%2F%28256%29+
+x+=+%289+%2B-+2551+%29%2F%28256%29+

so we have
+x+=+%289+%2B+2551+%29%2F%28256%29+ or +x+=+%289+-+2551+%29%2F%28256%29+

but the second will give a negative answer and seeing as how we are talking about years, this makes no sense. So concentrate on the first answer:

+x+=+%289+%2B+2551+%29%2F%28256%29+
+x+=+%282560%29%2F%28256%29+
x = 10

So 10 years after 1997 --> 2007.

cheers
Jon.